Question

. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1? (g= 9.8 m s–2)

Answer 1

As given Height of the tower = 300m  , Intial velocity of stone, u = 0 m/s , , Stone will free fall under the effect of gravity, so acceleration = g

let us consider, time taken by stone to reach the pond, t1 = ?

h = ut1 + (1/2)gt12

300 = 0 + (0.5 *x 9.8 x t12)

t12 = 61.22

t1 = 7.82 s

Now let us consider time taken by sound to reach the top of tower is t2.

t2 = h/v = 300/340 = 0.88 s

Total time, t = t1 + t2 = 7.82 + 0.88 = 8.7 s

Answer 2

Height = s = 300m
g = 9.8 m/s²
initial velocity = 0
speed of sound in air = 340m/s
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using newtons second equation of motion

s =  ut + 1/2 gt²
300 = 0 + 1/2 ( 9.8 ) t

300= 4.9t

t² =61.22

t1 = 7.82s
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for sound

we know that velocity = distance /time

therefore 

time = distance / velocity of sound in air

= 300/340

t2 =0.88 s
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for the splash to be heard we need to add the time taked to reach the water and time taken by the sound to reach us

therefore

T = t1 + t2

= 7.82 + 0.88

=8.7s
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after 8.7 s we will hear the splash